# Kerodon

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### 10.2.4 Universal Quotient Morphisms

Let $\operatorname{\mathcal{C}}$ be an $\infty$-category. In §10.2.2, we observed that the collection of quotient morphisms in $\operatorname{\mathcal{C}}$ can exhibit some bad behavior: they need not be closed under composition (Exercise 10.2.2.15) or under the formation of pullbacks (Exercise 10.2.2.16). These deficiencies can be remedied by adopting a more restrictive definition.

Definition 10.2.4.1 (Universal Quotient Morphisms). Let $\operatorname{\mathcal{C}}$ be an $\infty$-category, let $f: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$, and let $\operatorname{\mathcal{C}}^{0}_{ / Y } \subseteq \operatorname{\mathcal{C}}_{/Y}$ be the sieve generated by $f$ (see Example 10.2.1.19). We say that $f$ is a universal quotient morphism if the sieve $\operatorname{\mathcal{C}}^{0}_{/X}$ is dense (in the sense of Definition 10.2.1.26).

Remark 10.2.4.2. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $f: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$, and let $\operatorname{\mathcal{C}}^{0}_{/Y} \subseteq \operatorname{\mathcal{C}}_{/Y}$ be the sieve generated by $f$. Then $f$ is a quotient morphism if and only if the forgetful functor $\operatorname{\mathcal{C}}_{/Y} \rightarrow \operatorname{\mathcal{C}}$ is left Kan extended from $\operatorname{\mathcal{C}}^{0}_{/Y}$ at the object $\operatorname{id}_{Y}: Y \rightarrow Y$ (see Remark 10.2.1.32). In particular, if $f$ is a universal quotient morphism, then $f$ is a quotient morphism. Beware that the converse is false in general (see Example 10.2.4.11).

Example 10.2.4.3. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category, let $f: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$, and let $\operatorname{\mathcal{C}}^{0}_{/Y} \subseteq \operatorname{\mathcal{C}}_{/Y}$ be the sieve generated by $f$. If $f$ admits a right homotopy inverse $s: Y \rightarrow X$, then the sieve $\operatorname{\mathcal{C}}^{0}_{/Y}$ coincides with $\operatorname{\mathcal{C}}_{/Y}$, and is therefore dense. It follows that $f$ is a universal quotient morphism. In particular, every isomorphism is a universal quotient morphism.

Remark 10.2.4.4. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category, let $Y$ be an object of $\operatorname{\mathcal{C}}$, and let $\operatorname{\mathcal{C}}^{0}_{/Y} \subseteq \operatorname{\mathcal{C}}_{/Y}$ be a sieve on $Y$. If $\operatorname{\mathcal{C}}^{0}_{/Y}$ contains a universal quotient morphism $f: X \twoheadrightarrow Y$, then it is dense. See Remark 10.2.1.31.

Remark 10.2.4.5. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category containing a $2$-simplex

$\xymatrix@R =50pt@C=50pt{ & Y \ar [dr]^{g} & \\ X \ar [ur]^{f} \ar [rr]^{h} & & Z. }$

If $h$ is a universal quotient morphism, then $g$ is also a universal quotient morphism. This is a special case of Remark 10.2.4.4.

Proposition 10.2.4.6. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category containing a pullback diagram

10.26
$$\begin{gathered}\label{equation:pullback-of-quotient} \xymatrix@R =50pt@C=50pt{ X' \ar [d]^{f'} \ar [r] & X \ar [d]^{f} \\ Y' \ar [r]^-{u} & Y. } \end{gathered}$$

If $f$ is a universal quotient morphism, then $f'$ is also a universal quotient morphism.

Proof. Let $\operatorname{\mathcal{C}}^{0}_{/Y} \subseteq \operatorname{\mathcal{C}}_{/Y}$ be the sieve generated by $f$. Our assumption that $f$ is a universal quotient morphism guarantees that $\operatorname{\mathcal{C}}^{0}_{/Y}$ is a dense sieve on $Y$. Applying Proposition 10.2.1.33, we deduce that the pullback $u^{\ast } \operatorname{\mathcal{C}}^{0}_{/Y}$ is a dense sieve on $Y'$. Since (10.26) is a pullback square, the sieve $u^{\ast } \operatorname{\mathcal{C}}^{0}_{/Y'}$ is generated by $f'$, so that $f'$ is also a universal quotient morphism. $\square$

The terminology of Definition 10.2.4.1 is motivated by the following result:

Corollary 10.2.4.7. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category which admits fiber products and let $f: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$. The following conditions are equivalent:

$(1)$

The morphism $f$ is a universal quotient morphism.

$(2)$

For every pullback diagram

$\xymatrix@R =50pt@C=50pt{ X' \ar [d]^{f'} \ar [r] & X \ar [d]^{f} \\ Y' \ar [r] & Y }$

of $\operatorname{\mathcal{C}}$, the morphism $f'$ is a universal quotient morphism.

$(3)$

For every pullback diagram

$\xymatrix@R =50pt@C=50pt{ X' \ar [d]^{f'} \ar [r] & X \ar [d]^{f} \\ Y' \ar [r] & Y }$

of $\operatorname{\mathcal{C}}$, the morphism $f'$ is a quotient morphism.

Proof. The implication $(1) \Rightarrow (2)$ follows from Proposition 10.2.4.6, the implication $(2) \Rightarrow (3)$ from Remark 10.2.4.2, and the implication $(3) \Rightarrow (1)$ from the criterion of Remark 10.2.1.32 (together with Example 10.2.1.25). $\square$

Corollary 10.2.4.8. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category which admits fiber products. The following conditions are equivalent:

$(1)$

Every quotient morphism in $\operatorname{\mathcal{C}}$ is a universal quotient morphism.

$(2)$

The collection of quotient morphisms in $\operatorname{\mathcal{C}}$ is closed under pullbacks. That is, for every pullback diagram

$\xymatrix@R =50pt@C=50pt{ X' \ar [d]^{f'} \ar [r] & X \ar [d]^{f} \\ Y' \ar [r] & Y }$

where $f$ is a quotient morphism, $f'$ is also a quotient morphism.

Corollary 10.2.4.9. Let $X$ and $Y$ be sets, and let $f: X \rightarrow Y$ be a function. The following conditions are equivalent:

$(1)$

The function $f$ is a universal quotient morphism in the category of sets.

$(2)$

The function $f$ is a quotient morphism in the category of sets.

$(3)$

The function $f$ is surjective.

Proof. The implication $(1) \Rightarrow (2)$ follows from Remark 10.2.4.2 and the equivalence $(2) \Leftrightarrow (3)$ follows from Example 10.2.2.8. Since the collection of surjections is closed under pullbacks, Corollary 10.2.4.7 guarantees that $(3) \Rightarrow (1)$. $\square$

Corollary 10.2.4.10. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category which admits pullbacks, and suppose that geometric realizations in $\operatorname{\mathcal{C}}$ are universal (see Definition ). Then every quotient morphism in $\operatorname{\mathcal{C}}$ is a universal quotient morphism.

Example 10.2.4.11. Let $\operatorname{\mathcal{C}}$ be (the nerve of) the category of partially ordered sets. Then Exercise 10.2.2.16 supplies an example of a quotient morphism $f: Q \twoheadrightarrow [2]$ in $\operatorname{\mathcal{C}}$ which is not a universal quotient morphism.

Proposition 10.2.4.12. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category containing a $2$-simplex

$\xymatrix@R =50pt@C=50pt{ & Y \ar [dr]^{g} & \\ X \ar [ur]^{f} \ar [rr]^{h} & & Z. }$

If $f$ and $g$ are universal quotient morphisms, then $h$ is also a universal quotient morphism.

Proof. Let $\operatorname{\mathcal{C}}^{0}_{/Z}$ and $\operatorname{\mathcal{C}}^{1}_{/Z}$ be the sieves generated by $g$ and $h$, respectively. By assumption, the sieve $\operatorname{\mathcal{C}}^{0}_{/Z}$ is dense, and we wish to show that $\operatorname{\mathcal{C}}^{1}_{/Z}$ is also dense. By virtue of Proposition 10.2.1.34, it will suffice to show that for every morphism $u: Z' \rightarrow Z$ which belongs to $\operatorname{\mathcal{C}}^{0}_{/Z}$, the pullback $u^{\ast } \operatorname{\mathcal{C}}^{1}_{/Z'}$ is a dense sieve on $Z'$. Using Proposition 10.2.1.33, we can reduce to the special case where $u$ is the morphism $g: Y \twoheadrightarrow Z$. In this case, the pullback sieve $u^{\ast }( \operatorname{\mathcal{C}}^{1}_{/Y}) \subseteq \operatorname{\mathcal{C}}_{/Y}$ contains the universal quotient morphism $f: X \twoheadrightarrow Y$, and is therefore dense (Remark 10.2.4.4). $\square$

Variant 10.2.4.13. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category containing a $2$-simplex

$\xymatrix@R =50pt@C=50pt{ & Y \ar [dr]^{g} & \\ X \ar [ur]^{f} \ar [rr]^{h} & & Z. }$

If $f$ is a universal quotient morphism and $g$ is a quotient morphism, then $h$ is a quotient morphism.

Proof. Let $\operatorname{\mathcal{C}}^{0}_{/Z}$ and $\operatorname{\mathcal{C}}^{1}_{/Z}$ be the sieves on $X$ generated by $g$ and $h$, respectively. Our assumption that $g$ is a quotient morphism guarantees that the functor

$\overline{Q}: (\operatorname{\mathcal{C}}^{0}_{/Z})^{\triangleright } \rightarrow (\operatorname{\mathcal{C}}_{/Z})^{\triangleright } \rightarrow \operatorname{\mathcal{C}}$

is a colimit diagram in the $\infty$-category $\operatorname{\mathcal{C}}$, and we wish to show that the restriction $\overline{Q}|_{ (\operatorname{\mathcal{C}}^{1}_{/Z})^{\triangleright }}$ is also a colimit diagram. By virtue of Corollary 7.3.8.2, it will suffice to show that the restriction $Q = \overline{Q}|_{ \operatorname{\mathcal{C}}^{0}_{/Z} }$ is left Kan extended from the full subcategory $\operatorname{\mathcal{C}}^{1}_{/Z}$. Fix a morphism $u: Z' \rightarrow Z$ which belongs to the sieve $\operatorname{\mathcal{C}}^{0}_{/Z}$; we wish to show that $Q$ is left Kan extended from $\operatorname{\mathcal{C}}^{1}_{/Z}$ at $u$. In fact, we will prove a slightly stronger assertion: the pullback $u^{\ast }( \operatorname{\mathcal{C}}^{1}_{/Z} )$ is a dense sieve on $Z'$. Using Proposition 10.2.1.33, we are reduced to proving this in the special case where $u$ is the morphism $g: Y \rightarrow Z$. In this case, the sieve $u^{\ast }( \operatorname{\mathcal{C}}^{1}_{/Z} )$ contains the quotient morphism $f$, and is therefore dense by virtue of Remark 10.2.4.4. $\square$

Corollary 10.2.4.14. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category. Then the collection of universal quotient morphisms of $\operatorname{\mathcal{C}}$ is closed under retracts (in the $\infty$-category $\operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}})$).

Proof. Let $f: X \twoheadrightarrow Y$ be a universal quotient morphism in $\operatorname{\mathcal{C}}$ and let $f': X' \rightarrow Y'$ be a retract of $f$, so that we have a commutative diagram

$\xymatrix@R =50pt@C=50pt{ X' \ar [d]^{f'} \ar [r] & X \ar [d]^{f} \ar [r]^-{r_{X}} & X' \ar [d]^{f'} \\ Y' \ar [r] & Y \ar [r]^-{ r_{Y} } & Y' }$

where the vertical compositions are homotopic to the identity. We wish to show that $f'$ is also a universal quotient morphism. By virtue of Remark 10.2.4.5, it will suffice to show that the composition $(f' \circ r_{X}): X \rightarrow Y'$ is a universal quotient morphism. Using the commutativity of the diagram, we can write $f' \circ r_{X}$ as a composition of $r_{Y}$ with $f$. Since $f$ is a universal quotient morphism by assumption and $r_{Y}$ is a universal quotient morphism by virtue of Example 10.2.4.3, the desired result follows from Proposition 10.2.4.12. $\square$

Proposition 10.2.4.15. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category, let $q: K \rightarrow \operatorname{\mathcal{C}}$ be a diagram, and let $\widetilde{f}: \widetilde{X} \rightarrow \widetilde{Y}$ be a morphism in the $\infty$-category $\operatorname{\mathcal{C}}_{/q}$ having image $f: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$. If $f$ is a universal quotient morphism in $\operatorname{\mathcal{C}}$, then $\widetilde{f}$ is a universal quotient morphism in $\operatorname{\mathcal{C}}_{/q}$.

Proof. Set $\widetilde{\operatorname{\mathcal{C}}} = \operatorname{\mathcal{C}}_{/q}$, so that we have a commutative diagram of forgetful functors

$\xymatrix@R =50pt@C=50pt{ \widetilde{\operatorname{\mathcal{C}}}_{ / \widetilde{Y} } \ar [d]^{ \widetilde{U} } \ar [r]^-{V'} & \operatorname{\mathcal{C}}_{/Y} \ar [d]^{U} \\ \widetilde{\operatorname{\mathcal{C}}} \ar [r]^-{V} & \operatorname{\mathcal{C}}. }$

Let $\operatorname{\mathcal{C}}^{0}_{ / Y } \subseteq \operatorname{\mathcal{C}}_{/Y}$ denote the sieve generated by $f$. Since $f$ is a universal quotient morphism, the functor $U$ is left Kan extended from $\operatorname{\mathcal{C}}^{0}_{/Y}$. Note that $V$ is a right fibration (Proposition 4.3.6.1), so that $V'$ is a trivial Kan fibration (Corollary 4.3.7.13). In particular, $V'$ is a right fibration, so that the functor $U \circ V' = V \circ \widetilde{U}$ is left Kan extended from the subcategory $\widetilde{\operatorname{\mathcal{C}}}^{0}_{ / \widetilde{Y} } = V'^{-1} \operatorname{\mathcal{C}}^{0}_{/Y}$ (Corollary 7.3.8.5). Since the functor $V$ is conservative and creates colimits (Proposition 7.1.3.19), it follows that $\widetilde{U}$ is also left Kan extended from $\widetilde{\operatorname{\mathcal{C}}}^{0}_{ / \widetilde{Y} }$. We conclude by observing that $\widetilde{\operatorname{\mathcal{C}}}^{0}_{ / \widetilde{Y} }$ is the sieve generated by $\widetilde{f}$, so that $\widetilde{f}$ is a universal quotient morphism in $\widetilde{\operatorname{\mathcal{C}}}$. $\square$

We close this section by characterizing (universal) quotient morphisms in the $\infty$-category $\operatorname{\mathcal{S}}$ of spaces.

Lemma 10.2.4.16. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category, let $\operatorname{\mathcal{C}}' \subseteq \operatorname{\mathcal{C}}$ be a dense full subcategory, and let $f: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$. Suppose that, for every object $C \in \operatorname{\mathcal{C}}'$, postcomposition with $[f]$ induces a surjection $\operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{\mathcal{C}}}}( C, X ) \rightarrow \operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{\mathcal{C}}}}( C, Y )$. Then $f$ is a universal quotient morphism.

Proof. Our assumption that $\operatorname{\mathcal{C}}'$ is dense guarantees that the identity functor $\operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}$ is left Kan extended from $\operatorname{\mathcal{C}}'$. Let $U: \operatorname{\mathcal{C}}_{/Y} \rightarrow \operatorname{\mathcal{C}}$ be the projection map and let $\operatorname{\mathcal{C}}'_{/Y} \subseteq \operatorname{\mathcal{C}}_{/Y}$ denote the inverse image of $\operatorname{\mathcal{C}}'$. Since $U$ is a right fibration (Proposition 4.3.6.1), the functor $U$ is left Kan extended from $\operatorname{\mathcal{C}}'_{/Y}$. Let $\operatorname{\mathcal{C}}^{0}_{/Y} \subseteq \operatorname{\mathcal{C}}_{/Y}$ denote the sieve generated by $f$. Our hypothesis guarantees that $\operatorname{\mathcal{C}}^{0}_{/Y}$ contains $\operatorname{\mathcal{C}}'_{/Y}$. Applying Corollary 7.3.8.8, we conclude that $U$ is left Kan extended from $\operatorname{\mathcal{C}}^{0}_{/Y}$: that is, $f$ is a universal quotient morphism. $\square$

Proposition 10.2.4.17. Let $f: X \rightarrow Y$ be a map of Kan complexes. The following conditions are equivalent:

$(1)$

The map $f$ is a universal quotient morphism in the $\infty$-category $\operatorname{\mathcal{S}}$ (Definition 10.2.4.1).

$(2)$

The map $f$ is a quotient morphism in the $\infty$-category $\operatorname{\mathcal{S}}$ (Definition 10.2.2.1).

$(3)$

The map $f$ is $0$-connective: that is, it induces a surjection $\pi _0(f): \pi _0(X) \rightarrow \pi _0(Y)$.

Proof. The implication $(1) \Rightarrow (2)$ is a special case of Corollary 10.2.4.10. We next show that $(2)$ implies $(3)$. Assume that $f$ is a quotient morphism, so that $Y$ can be identified with the geometric realization of the Čech nerve $\operatorname{\check{C}}(X/Y)_{\bullet }$ in the $\infty$-category $\operatorname{\mathcal{S}}$ (Proposition 10.2.2.4). Note that the functor

$\operatorname{\mathcal{S}}\rightarrow \operatorname{N}_{\bullet }( \operatorname{Set}) \quad \quad S \mapsto \pi _0(S)$

preserves the formation of geometric realizations (since it is left adjoint to the inclusion functor). It follows that $\pi _0(Y)$ can be identified with the geometric realization of $\pi _0( \operatorname{\check{C}}(X/Y)_{\bullet } )$ in the category of sets: that is, with the coequalizer of the projection maps $\pi _0( X \times _{Y} X ) \rightrightarrows \pi _0(X)$ (see Corollary 10.1.2.12). In particular, the tautological map $\pi _0(X) \rightarrow \pi _0(Y)$ is surjective.

We now show that $(3)$ implies $(1)$. Assume that condition $(3)$ is satisfied. For every contractible Kan complex $C$, the composition map $\operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{\mathcal{C}}}}( C, X ) \xrightarrow { [f] \circ } \operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{\mathcal{C}}}}( C, Y )$ can be identified with $\pi _0(f)$, and is therefore surjective. Since the contractible Kan complexes span a dense subcategory of $\operatorname{\mathcal{S}}$ (Example 8.4.2.3), Lemma 10.2.4.16 implies that $f$ is a universal quotient morphism. $\square$